The Laplace transform from scratch

Here's a natural way to derive the Laplace transform via taking the concept of a taylor series and making it continuous. We will, of course, start with a taylor series whose coefficients are given by a(n)a(n)

0a(n)xnx<1 \qquad \qquad \displaystyle\sum_0^{\infty} a(n) x^n \qquad x<1

Then, make this continuous by replacing the sum by an integral

0a(n)xndnx<1 \qquad \qquad \displaystyle\int_0^{\infty} a(n) x^n dn \qquad x<1

We can change xnx^n to enlnxe^n \ln x and make the substitution s=lnxs =- \ln x to give us

0a(n)esndns>0 \qquad \qquad \displaystyle\int_0^{\infty} a(n) e^{-sn} dn \qquad s > 0

Which is just the definition for the Laplace transform of a(n)a(n), L(a)(s)\mathcal{L}(a)(s).

Note by Levi Walker
1 year, 7 months ago

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Krishna Karthik - 1 year, 3 months ago

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